Or maybe we can put it this way: if and does it follow that ?
Or maybe we can put it this way: if
and does it follow that ?
Not in general. You need the extra assumption that are never zero.
The proof is easy: Choose any natural number . Since there are integers such that . So ; Thus is differentiable as a product of differentiable functions. Note that the exponents may be negative, and so the assumption is essential.
I have another way to show this.
If for any point and then thus . Because the composition of infinitely differenciable functions is infinitely differenciable. Note, that is not differenciable at zero, which is why we need the condition that the function is never zero. However, I cannot come up with a conterexample.
This does not work with .
Ah! So my thought that we may construct the function f as is at least not out of the ballpark. (This, of course, gives the same limitation that at any point in its domain.) But this can't be the whole story because if we have
and
the function is, in fact, .
How might we get around this? (Or am I barking up the wrong tree again? )
-Dan